Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 7 - Section 7.3 - Simplifying Radicals, the Distance Formula, and Circles - 7.3 Exercises - Page 461: 144

Answer

Graph of $ (x-1)^2+(y+3)^2=16 $

Work Step by Step

Using $(x-h)^2+(y-k)^2=r^2$ or the Center-Radius Form, the given equation, $ (x-1)^2+(y+3)^2=16 ,$ is equivalent to \begin{align*}\require{cancel} (x-1)^2+(y-(-3))^2=4^2 .\end{align*} Since the center of a circle is given by $(h,k),$ then the center of the circle with the equation above is $ (1,-3) $. Since the radius is given by $r$, then the radius of the circle above is $ 4 $. Moving $r$ units to the right of the center results to the point $ (5,-3) $. Moving $r$ units above the center results to the point $ (1,1) $. Moving $r$ units to the left of the center results to the point $ (-3,-3) $. Moving $r$ units below the center results to the point $ (1,-7) $. Connecting these points give the graph of the circle defined by the equation $ (x-1)^2+(y+3)^2=16 $.
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